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concepts/em/u9/e9-4-equipotentials.tex

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+\subsection{Equipotentials and Energy Conservation for Moving Charges}
+
+This subsection explains how equipotential curves or surfaces encode the direction of $\vec{E}$ and how potential differences determine changes in kinetic and electric potential energy for moving charges.
+
+\dfn{Equipotentials and the energy-change relation}{Let $V(\vec{r})$ denote the electric potential at position vector $\vec{r}$. An \emph{equipotential} is a curve in two dimensions or a surface in three dimensions on which the potential has one constant value:
+\[
+V(\vec{r})=\text{constant}.
+\]
+If a charge $q$ moves from point $A$ to point $B$, define
+\[
+\Delta V=V_B-V_A
+\]
+and
+\[
+\Delta U=U_B-U_A.
+\]
+Then the change in electric potential energy is
+\[
+\Delta U=q\Delta V.
+\]
+Thus potential difference tells how the electric potential energy of a chosen charge changes between two points.}
+
+\thm{Equipotentials, no work, and kinetic-energy change}{Let $d\vec{\ell}$ denote an infinitesimal displacement in an electrostatic field $\vec{E}$. Then
+\[
+dV=-\vec{E}\cdot d\vec{\ell}.
+\]
+If the displacement is along an equipotential, then $dV=0$, so
+\[
+\vec{E}\cdot d\vec{\ell}=0.
+\]
+Therefore the electric field is perpendicular to an equipotential, and the electric force does no work on a charge moved along an equipotential:
+\[
+W_{\mathrm{elec}}=q\int \vec{E}\cdot d\vec{\ell}=0.
+\]
+For any motion of a charge $q$ from $A$ to $B$ in electrostatics,
+\[
+\Delta U=q\Delta V.
+\]
+If only the electric force does work, conservation of mechanical energy gives
+\[
+K_i+U_i=K_f+U_f,
+\]
+so
+\[
+\Delta K=K_f-K_i=-\Delta U=-q\Delta V.
+\]
+Hence a charge speeds up when its electric potential energy decreases.}
+
+\ex{Illustrative example}{Points $A$ and $B$ lie on the same equipotential,
+\[
+V_A=V_B=120\,\mathrm{V}.
+\]
+Let a proton of charge
+\[
+q=+e=+1.60\times 10^{-19}\,\mathrm{C}
+\]
+move from $A$ to $B$.
+
+Because the two points are on the same equipotential,
+\[
+\Delta V=V_B-V_A=0.
+\]
+So the change in electric potential energy is
+\[
+\Delta U=q\Delta V=0,
+\]
+and the work done by the electric field is also zero. The field may still be present, but along that displacement it is perpendicular to the motion.}
+
+\nt{Be careful to distinguish \emph{lower potential} from \emph{lower potential energy}. Since $\Delta U=q\Delta V$, a positive charge has lower potential energy at lower potential, but a negative charge has lower potential energy at \emph{higher} potential. Thus a positive charge released from rest tends to speed up toward lower $V$, whereas an electron released from rest tends to speed up toward higher $V$. In both cases the rule is the same: the charge moves spontaneously in the direction that makes $U$ decrease and $K$ increase.}
+
+\qs{Worked AP-style problem}{Two large parallel plates create a uniform electrostatic region. Let point $A$ be near the negative plate and point $B$ be near the positive plate. The potentials are
+\[
+V_A=100\,\mathrm{V},
+\qquad
+V_B=400\,\mathrm{V}.
+\]
+An electron is released from rest at point $A$ and moves to point $B$. Let the electron charge be
+\[
+q=-1.60\times 10^{-19}\,\mathrm{C}
+\]
+and the electron mass be
+\[
+m_e=9.11\times 10^{-31}\,\mathrm{kg}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the potential difference $\Delta V=V_B-V_A$,
+\item the change in electric potential energy $\Delta U$,
+\item the change in kinetic energy $\Delta K$, and
+\item the electron's speed at point $B$.
+\end{enumerate}}
+
+\sol First compute the potential difference:
+\[
+\Delta V=V_B-V_A=400\,\mathrm{V}-100\,\mathrm{V}=300\,\mathrm{V}.
+\]
+
+For part (b), use the relation
+\[
+\Delta U=q\Delta V.
+\]
+Substitute the electron charge and the potential difference:
+\[
+\Delta U=(-1.60\times 10^{-19}\,\mathrm{C})(300\,\mathrm{V}).
+\]
+Since $1\,\mathrm{V}=1\,\mathrm{J/C}$,
+\[
+\Delta U=-4.80\times 10^{-17}\,\mathrm{J}.
+\]
+
+For part (c), only the electric force does work, so
+\[
+\Delta K=-\Delta U.
+\]
+Therefore,
+\[
+\Delta K=+4.80\times 10^{-17}\,\mathrm{J}.
+\]
+
+This sign makes sense. The electron moves toward higher potential, but because its charge is negative, that motion lowers its electric potential energy and increases its kinetic energy.
+
+For part (d), the electron starts from rest, so
+\[
+K_i=0.
+\]
+Thus
+\[
+K_f=\Delta K=4.80\times 10^{-17}\,\mathrm{J}.
+\]
+Use the kinetic-energy formula
+\[
+K_f=\frac12 m_e v^2.
+\]
+Solve for the speed $v$:
+\[
+v=\sqrt{\frac{2K_f}{m_e}}.
+\]
+Substitute the values:
+\[
+v=\sqrt{\frac{2(4.80\times 10^{-17}\,\mathrm{J})}{9.11\times 10^{-31}\,\mathrm{kg}}}.
+\]
+This gives
+\[
+v=1.03\times 10^7\,\mathrm{m/s}.
+\]
+
+Therefore,
+\[
+\Delta V=300\,\mathrm{V},
+\qquad
+\Delta U=-4.80\times 10^{-17}\,\mathrm{J},
+\]
+\[
+\Delta K=+4.80\times 10^{-17}\,\mathrm{J},
+\qquad
+v=1.03\times 10^7\,\mathrm{m/s}.
+\]